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FOCUS 2 MOLECULES · 2018-09-16 · Focus 2 Molecules 85 2C.16 2C.18 (a) The formal charge...
Transcript of FOCUS 2 MOLECULES · 2018-09-16 · Focus 2 Molecules 85 2C.16 2C.18 (a) The formal charge...
FOCUS 2
MOLECULES
2A.2 (a) 5; (b) 8; (c) 12; (d) 7
2A.4 (a) [Ar] 3d1; (b) [Ar] 3d5; (c) [Kr] 4d10; (d) [Xe]
2A.6 (a) [Kr]; (b) [Xe] 4f145d5; (c) [Xe]; (d) [Ar]
2A.8 (a) Ca: [Ar] 4s2, Ti2+: [Ar] 3d2; V3+: [Ar] 3d2 In the d block, the energies
of the n-1 d-orbitals lie below those of the ns-orbitals. Therefore, when V
and Ti form ions they lose their 3s electrons before losing their 4d
electrons. (b) Ca: no unpaired electrons; Ti2+: two unpaired electrons; V3+:
two unpaired electrons. (c) Ti3+: [Ar] 3d1 no neutral atom has this electron
configuration.
2A.10 (a) Au3+; (b) Os3+; (c) I3+; (d) As3+
2A.12 (a) Cr2+; (b) Ag2+; (c) Zn2+; (d) Pb2+
2A.14 (a) 4d; (b) 3p; (c) 5p; (d) 4s
2A.16 (a) +1; (b) –2; (c) +2; (d) –3; (e) –1
2A.18 (a) 2; (b) 5; (c) 3; (d) 6
2A.20 (a) [Ar]; no unpaired electrons; (b) [Ar]3d7; three unpaired electrons;
(c) [Kr]; no unpaired electrons; (d) [Kr]; no unpaired electrons
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78 Focus 2 Molecules
2A.22 (a) [Ar]3d5; five unpaired electrons; (b) [Xe]4f145d106s2; no unpaired
electrons; (c) [Ne]; no unpaired electrons; (d) [Xe]; no unpaired
electrons.
2A.24 (a) MnTe; (b) Ba3As2; (c) Si3N4; (d) Li3Bi; (e) ZrCl4
2A. 26
2A.28 The coulombic attraction is inversely proportional to the distance between
the two oppositely charged ions (Equation 1) so the ions with the shorter
radii will give the greater coulombic attraction. The answer is therefore
(c) Mg2 , O2 .
2A.30 The Br ion is smaller than the I ion (196 vs 220 pm). Because the
lattice energy is related to the coulombic attraction between the ions, it
will be inversely proportional to the distance between the ions (see
Equation 2). Hence the larger iodide ion will have the lower lattice energy
for a given cation.
2B.2
O ClCl
H
GeH H
H(a)Br P Br
Br Cl
GaCl Cl(c)(b) (d)
Focus 2 Molecules 79
2B.4
2B.6
2B.8 The structure has a total of 32 electrons; of these, 14 are accounted for by
the chlorines (2 Cl’s 7 valence electrons each); and 12 are accounted for
by the oxygens (2 O’s 6 valence electrons each). This leaves 6 electrons
unaccounted for; these must come from E. Therefore, E must be a
member of the oxygen family, and since it is a fourth period element, E
must be selenium (Se)
2B.10
(a)
O OClO O
O
COH
NO O2
(b) (c) (d)
C N
C NCa2+ O S O
O
O
2
K+ K+
H N H
H
H
O I O
O
(a) (b)
(c)
80 Focus 2 Molecules
2B.12
2B.14
2B.16
2B.18 There are four possible resonance structures for Se42 (the seleniums have
been numbered for clarity):
SeSe
Se Se
1 4
32
1 4
32
Se
Se Se
Se
Se Se
SeSe
Se
SeSe
Se
1 4
32
1 4
32
2+ 2+
2+2+
C
N
N NH
H
H
H
HH
C
N
N NH
H
H
H
HH
C
N
N NH
H
H
H
HH
Focus 2 Molecules 81
2B.20
2B.22 The three possible Lewis structures for NOF are:
The second structure is the most likely since all atoms have zero formal
charge.
2B.24
2C.2 Radicals are species with an unpaired electron; therefore, only (a) and
(d) are radicals since they have an odd number of electrons while (b)
and (c) have an even number of electrons, which allows Lewis
structures to be drawn with all electrons paired.
H
CH H
-1
All H's are 0
Br OO-1 +1 -1
O P O
O-1
-1
-10
3(a) (b) (c)
N O F0 0
O N F N F O0 -1 +2 -1-1 +1 0
O As O
O
O
-1-1
-1-1
+1
3-
O As O
O
O
-10
-1-1
0
3-
O As O
O
O
00
00
3-
lower energy
-3
O I O
O
O
-1
-1
-1
-1
+3O I O
O
O
0
0
-1
0
0O I O
O
O
0
-1
-1
0
lower energy
+1
N C N0-1 -1
N C N0-2 0
lower energy
2 2
(a)
(b)
(c)
82 Focus 2 Molecules
2C.4 (a) The formate ion has two resonance structures, each having an
oxygen with a formal charge of –1:
(b) The hydrogen phosphite ion has one Lewis structure that obeys
the octet rule, the first structure shown below. Including one double
bond to oxygen lowers the formal charge of P from +1 to 0. Three
resonance forms include this contribution.
(c) The bromate ion has one Lewis structure that obeys the octet
rule, the first structure shown below. Including double bonds to two
of the oxygens lowers the formal charge of Br from +2 to 0. Three
resonance forms include this contribution.
(d) The selenate ion has one Lewis structure that obeys the octet rule:
C
O
H O0 C
O
H O-1
0-1
Focus 2 Molecules 83
Including double bonds to two of the oxygens lowers the formal
charge of Se from +2 to 0. Six resonance forms include this
contribution.
2C.6
2C.8
(a) (b)Cl P Cl
Cl P Cl
+
Cl
Cl(c) (d)
ClP Cl
Cl
Cl
Cl
Cl
ClP
Cl
Cl
Cl
ClCl
P has 3 bonding pairsand 1 lone pair
P has 5 bonding pairsand no lone pair
P has 4 bonding pairsand no lone pair
P has 6 bonding pairsand no lone pair
84 Focus 2 Molecules
ICl Cl
10 electrons
SCl Cl
8 electronsO Si O
O
O
4-
FBr F
F
F
F
12 electrons
8 electrons
(c) (d)
(a) (b)
2C.10
]-
2C.12
2C.14 (a) In SF6, there are 12 electrons around the central sulfur:
(b) In BH3 there are only six electrons around the central B:
H B H
H
Focus 2 Molecules 85
2C.16
2C.18 (a) The formal charge distribution is similar for both structures. In the
first, the end nitrogen atom is 1 , the central N atom is +1, and O atom is
0. For the second structure, the end N atom is 0, the central N atom is still
+1, but the O atom is 1 . The second is preferred because it places the
negative formal charge on the most electronegative atom. (b) In the first
structure, there are three O atoms with formal charges of 1 and one O
atom with a formal charge of 0. The formal charge on the P atom is 0. In
the second structure, the P atom has a formal charge of 1 ; there are two
oxygen atoms with formal charges of 1 and two with formal charges of
0. The first structure is preferred because it places the negative formal
charge at the more electronegative atom, in this case O.
2D.2 As (2.18) < S (2.58) < I (2.66) < Br (2.96) < O (3.44). Generally
electronegativity increases as one goes from left to right across the
periodic table and as one goes from heavier to lighter elements within a
group.
2D.4 CCl4 would have bonds that are primarily covalent. The electronegativity
difference between C and Cl is smaller then between Ca or Cs and Cl,
making the C-Cl bond more covalent.
(b) O S O
O0
0
0
-2
lower energy
O S O
O-1
-1
-1
+1
(a) O S O-1 0+1
lower energy
O S O0 00
2-2-
86 Focus 2 Molecules
2D.6 (a) The N—H bond in NH3 would be more ionic; the electronegativity
difference between N and H (3.04 versus 2.20) is greater than between P
and H (2.19 versus 2.20). (b) N and O have similar electronegativities
(3.04 versus 3.44), which leads us to expect that the N—O bonds in NO2
would be fairly covalent. The electronegativity difference between S and
O is greater, so S—O bonds would be expected to be more ionic (2.58
versus 3.44). (c) Difference between SF6 and IF5 would be small
because S and I have very similar electronegativities (2.58 versus 2.66).
Because I has an electronegativity closer to that of F, IF5 can be expected
to have more covalent bond character, but probably only slightly more,
than SF6.
2D.8 (a) The electronegativity difference between Li and I (1.68) is
greater than that between Mg and I (1.35), so LiI should be more
soluble in water than MgI2. (b) The electronegativity difference
between Ca and S is 1.58, so while the difference is 2.44 between Ca
and O, CaO should be more water soluble than CaS.
2D.10 Cs K Mg2 Al3 : the smaller, more highly charged cations will be
the more polarizing. The ionic radii are 170 pm, 138 pm, 72 pm, and 53
pm, respectively.
2D.12 O2− < N3− < Cl− < Br−: the polarizability should increase as the ion gets
larger and less electronegative.
2D.14 (a) NO NO2 NO3
Bond order indicates the number of bonds between atoms. Examine the
type of bond between the nitrogen and oxygen atoms by drawing Lewis
structures. In NO it is a double bond (bond order of 2), in NO2 it is an
average of a double and a single bond (bond order of 1.5) because of
Focus 2 Molecules 87
resonance, and in NO3− it is an average of two single bonds and one
double bond (bond order of 1.33) because of resonance.
(b) C2H2 > C2H4 > C2H6
Examine the type of bond between the two carbon atoms by drawing
Lewis structures. In C2H2 it is a triple bond (bond order of 3), in C2H4 it is
a double bond (bond order of 2), and in C2H6 it is a single bond (bond
order of 1).
(c) H2CO > CH3OH ~ CH3OCH3
Examine the type of bond between the carbon and oxygen atoms by
drawing Lewis structures. In H2CO it is a double bond (bond order of 2)
and in both CH3OH and CH3OCH3 it is a single bond (bond order of 1).
2D.16 Bond strength increases as bond order increases. Examine the type of bond
between the carbon and nitrogen atoms by drawing Lewis structures. In
NHCH2 it is a double bond (bond order of 2), in NH2CH3 it is a single
bond (bond order of 1), and in HCN it is a triple bond (bond order of 3).
Therefore, the CN bond in HCN will be the strongest.
2D.18 (a) The C—O bond in formaldehyde is a double bond, so the expected
bond length will be 67 pm (double-bond covalent radius of C) + 60 pm
(double-bond covalent radius of O) = 127 pm. The experimental value is
120.9 pm. (b) and (c) The C—O bonds in dimethyl ether and methanol
are single bonds. The sum of the covalent single bond radii is 77 + 66 pm
= 143 pm. The experimental value in methanol is 142.7 pm. (d) The
C—S bond in methanethiol is a single bond. The sum of the covalent
single-bond radii is thus 77 + 102 pm = 179 pm. The experimental value is
181 pm.
2D.20 (a) The covalent radius of N is 75 pm, so the N—N single bond in
hydrazine is expected to be ca. 150 pm. The experimental value is 145 pm.
(b) The bond between nitrogen is a double bond; the sum of the covalent
88 Focus 2 Molecules
radii is 60 + 60 pm = 120 pm. The experimental value in methanol is 124
pm. (c) The lowest-energy Lewis structure for azide has a double bond
linking each terminal nitrogen to the central N; we would predict this
molecule to have two equal bond lengths of 60 + 60 pm = 120 pm.
Experimentally, azide has been shown to possess two equal-length N—N
bonds of 116 pm.
2E.2 (a) May have lone pairs
(b) Must have lone pairs
2E.4
2E.6
(a) The chlorite ion is angular (or bent). (b) All the oxygen atoms are
equivalent (resonance forms), so there should be only one O—Cl—O
bond angle; because of the presence of the two lone pairs of electrons on
the central chlorine, the bond angle is expected to be <109.5 .
2E.8
(a) The shape of the XeF4 molecule is square planar based upon an
octahedral arrangement of electrons about the Xe atom. (b) 90.
F
Se
F
F
F
FF
F
I
F
F
F
FF
F
(a) (b)
octahedral pentagonal bipyramidal
OClO O Cl O
Focus 2 Molecules 89
2E.10
(a) Tetrahedral; (b) 109.5
2E.12
(a) The phosphorus atom in PF4 will have five pairs of electrons about it.
There will be four bonding pairs and one lone pair. The arrangement of
electron pairs will be trigonal bipyramidal with the lone pair occupying an
equatorial position in order to minimize e-e repulsions. The name of the
shape ignores the lone pairs, so the molecule is best described as having a
seesaw structure. AX4E (b) The number and types of lone pairs are the
same for [ICl4]as for PF4 . The structural arrangement of electron pairs
and name are the same as in (a). AX4E (c) As in (a), the central P atom
has five pairs of electrons about it, but this time they all are bonding pairs.
The arrangement of pairs is still trigonal bipyramidal, but this time the
name of the molecular shape is the same as the arrangement of electron
pairs. The molecule is therefore a trigonal bipyramid. AX5 (d) Xenon
tetrafluoride will have six pairs of electrons about the central atom, of
which two are lone pairs and four are bonding pairs. These pairs will adopt
an octahedral geometry, but because the name of the molecule ignores the
lone pairs, the structure will be called square planar. The lone pairs are
F
P
F
F
F
Cl
I
Cl
Cl
Cl
FXe
F
F F
FP F
F
F
F
(a) (b) (c) (d)
90 Focus 2 Molecules
placed opposite each other rather than adjacent, to minimize e-e repulsions
between the lone pairs. AX4E2
2E.14 The Lewis structures:
(a) SiCl4 has an AX4 VSEPR formula resulting in a molecular shape that
is tetrahedral with angles of 109.5. (b) PF5 has an AX5 VSEPR formula
resulting in a molecular shape of trigonal bipyramidal; F—P—F bond
angles are 90 and 120. (c) SBr2 is AX2E2, resulting in an angular
molecular shape with Br—S—Br bond angles of <109.5. (d) ICl2+ is
AX2E2, resulting in an angular molecular shape with Cl—I—Cl bond
angles of <109.5.
2E.16 The Lewis structures:
(a) PCl3F2 is trigonal bipyramidal with angles of 120, 90, and 180.
The most symmetrical structure is shown with all Cl atoms in equatorial
positions. Phosphorus pentahalide compounds with more than one type of
(a)
Si
Cl
Cl
Cl Cl
(b)
PF
F
F
FF
(c)
SBr Br
(d)
ICl Cl
FI
F
F
F
F
F
Sn
F
F
F
FF
2
ClP Cl
Cl
F
F
F Sn F
F
F(a) (b) (c)
O Xe O
O
O
(e)(d)
Focus 2 Molecules 91
halogen atom like this have several possible geometrical arrangements of
the halogen atoms. These different arrangements are known as isomers.
For this type of compound, the energy differences between the isomers are
low, so that the compounds exist as mixtures of isomers that are rapidly
interconverting. AX5 (or AX3 X 2 ). (b) SnF4 is tetrahedral with
F—Sn—F bond angles of 109.5. AX4. (c) SnF62 is octahedral with
F—Sn—F bond angles of 90 and 180. AX6. (d) IF5 is square pyramidal
with F—I—F angles of approximately 90 and 180. AX5E. (e) XeO4 is
tetrahedral with angles equal to 109.5. AX4
2E.18 (a) slightly less than 109.5; (b) slightly less than 109.5; (c) slightly
less than 120; (d) slightly less than 109.5°
2E.20 The Lewis structures are:
(a) No lone pairs on central atom; trigonal bipyramidal; 90o, 120o
(b) Two lone pairs on central atom; T-shaped; 90o, 180o
(c) No lone pairs on central atom; trigonal bipyramidal; 90o, 120o
(d) Two lone pairs on central atom; T-shaped; 90o, 180o
(e) One lone pair on central atom, trigonal pyramidal; 107o
2E.22 The Lewis structures:
BrP Br
Br
Br
Br
O Xe
F
F
F I
F
F
FS F
F
F
F
BrO
OO
(a) (b) (c)
(d) (e)
92 Focus 2 Molecules
(a) square pyramidal; (b) trigonal bipyramidal; (c) trigonal
bipyramidal; (d) octahedral
2E.24 The Lewis structures:
(a) tetrahedral; (b) trigonal bipyramidal; (c) trigonal bipyramidal
2E.26 The Lewis structures:
Molecules (a) and (d) will be polar; (b) and (c) will be nonpolar.
FCl F
F
F
F
FSb F
F
F
F
OI O
O
O
O
O
I
O
O
O
OO
3 5
(a) (b)
(c) (d)
O As O
O
O
3-
FI O
F
F
O
FS F
F
F
O
(a) (b) (c)
FAs F
F
F
F
O Si O
H Se H
F N F
F
(a) (b)
(c) (d)
Focus 2 Molecules 93
2E.28
All of the compounds are polar.
2E.30 (a) In 2 the C—H and C—F bond vectors oppose identical bonds on
opposite ends of the molecule; the individual dipole moments will cancel
so that 2 will be nonpolar. This is not true for either 1 or 3, which will
both be expected to be polar. (b) Assuming the C—F and C—H
polarities are the same in molecules 1 and 3, one can carry out a vector
addition of the individual dipole moments. It is perhaps easiest to look at
the resultant of the two F—C dipoles and the two C—H dipoles in each
molecule. The dipoles will sum as shown:
Addition of 2 C—F and 2 C—H dipoles in 1 and 3 give:
Net dipoles:
The net C—F and C—H dipoles reinforce in both these molecules, but
because the F—C—F and H—C—H angles in 3 are more acute, the
magnitude of the resultant will be slightly larger for 3.
(a)
C
H
H
H S
(b) (c)
H C
H
H
H N H
H
C
H
H
H O C
H
H
H
C C
F
F H
H
C C
H
F F
H1 3
C C
F
F H
H
C C
H
F F
H1 3
C C
F
F H
H
C C
H
F F
H1 3
94 Focus 2 Molecules
2F.2 (a) sp3 (b) sp3d (c) sp3d2 (d) sp
2F.4 (a) 1 bond. 1 bond (b) 2 bonds. 2 bonds
2F.6 (a) sp3d2 (b) sp3 (c) sp2 (d) sp2
2F.8 (a) sp (b) sp2 (c) sp2 (d) sp3
2F.10 (a) sp (b) sp3 (c) sp (d) sp2
2F.12 The Lewis structure of 2P is similar to that of 2N :
However, P is a larger atom than N, so the PP bond is longer, resulting
in poorer overlap and therefore making a weaker and more reactive bond
2F.14
The first two molecules have four groups around the central atom, leading
to tetrahedral dispositions of the bonds and lone pairs. XeO3 is of the
AX3E type and will be pyramidal, whereas XeO4 will be of the AX4 type
and will be tetrahedral. The XeO64 ion will be octahedral. The
hybridizations will be sp3, sp3, and sp3d2, respectively.
The Xe—O bonds should be longest in XeO62 because each of those
bonds should have a bond order of ca. (4 1 2 2)/6 1.5, whereas the
bond orders in XeO3 and XeO4 will be about 2. This agrees with
experiment: XeO3, 174 pm; XeO4, 176 pm; XeO62 , 186 pm.
N N P Pvs.
O Xe
O
O
O Xe O
O
O
O
Xe
O
O
O
OO
4
Focus 2 Molecules 95
2F.16 (a) In the NH2 molecule, there is a single lone pair of electrons on the
central N atom, and as a result, the hybridization about the N atom is sp2 .
In NH2 , there are two lone pairs of electrons on the central N atom and
the hybridization about this atom is best described as sp3. (b) The N 2px
orbital contributes to bonding in the NH2 molecule (all p-orbitals
participate in the hybrid orbitals), but the N 2px orbital does not contribute
to bonding in NH2 . In this molecule, the N 2px orbital lies normal to the
plane containing the bonds and the lone pair; that is, there is no overlap
between the N 2px and the orbitals responsible for bonding and the orbital
containing the lone pair of electrons.
2F.18 The Lewis structure of formamide:
Here, both the C and the O are sp2 hybridized while the N is sp3
hybridized. The H—C—O, H—C—N, and O—C—N bond angles are
each 120; the molecule has five sigma bonds (one each connecting the
H’s to the C and the N, one connecting the N to the C, and one O to the
C). Finally, there is one pi bond (between the C and the O).
2F.20 If we take the hybrid orbital to be N h1, then normalization requires that:
N h1 2 d 1
solving for N:
N2 h12 d N2 (s px py pz )2 d 1
Expanding this expression we find that the only terms that are not zero
are:
C N
O
HH
H
96 Focus 2 Molecules
N2 s2 px2 py
2 pz2 d N2 4 1
Therefore, N 14
12
and the normalized wavefunction is 12
h1
2F.22
2G.2 The MO energy diagrams for B2, B2 , and B2
are
(a) B2 bond order = ½ (2 +2 –2) = 1 Paramagnetic, 2 unpaired electrons
(b) B2 bond order = ½ (2 + 3 − 2) = 3/2 Paramagnetic, one unpaired
electron
(c) B2+ bond order = ½ (2 + 1 − 2) = 1/2 Paramagnetic, one unpaired
electron
2G.4 (a) (i) ( 2s )2 ( *2s )2 ( 2p )4 ( 2p )1
2p
2p
*
2p
2p*
2s*
2s
2p
2p
*
2p
2p*
2s*
2s
2p
2p
*
2p
2p*
2s*
2s
(a) (b) (c)
0
1
2
90 95 100 105 110 115 120
Focus 2 Molecules 97
(ii) ( 2s )2 (*2 s )2 ( 2p )4
(iii) ( 2s )2 (*2 s )2 ( 2p )4 ( 2p )2 (*2p )2
(b) (i) 2.5
(ii) 2
(iii) 2
(c) (i) and (iii) are paramagnetic
(d) (i)
(ii)
(iii)
2G.6 The charge on B2n is –1 and the bond order is 1.5
1bond order = (2 3 2) = 1.52
2G.8 B2 is the only other period 2 homonuclear diatomic molecule that is
expected to be paramagnetic because of the two unpaired electrons in the
p molecular orbitals.
2G.10 (a) In molecular orbital theory, ionic and covalent bonding are extremes
of the same phenomenon. According to molecular orbital theory, bonding
occurs when at least one orbital on each of two atoms combines to form a
bonding and antibonding set of molecular orbitals. If the orbitals on each
atom that are used to make the molecular orbital are similar in energy, the
resultant molecular orbitals will be composed almost equally of
contributions from the two atoms. The result is a covalent bond. On the
other hand, if the two orbitals that make up the molecular orbitals are quite
different in energy, the bond will be highly polarized. The more
electronegative atom will contribute more to the bonding orbital, and the
more electropositive atom will contribute more to the antibonding orbital.
(b) The electronegativity of an atom is reflected in the energy of its
98 Focus 2 Molecules
atomic orbitals. The more electronegative the atom, the lower the orbitals.
As the electronegativity difference becomes larger, the atomic orbitals on
the two atoms being combined to form the molecular orbitals become
farther apart in energy, so that the bond becomes more polarized.
2G.12 (a) O22 (14 valence electrons): ( 2s )
2(*2s )2( 2p )2( 2p )4(*2p )4 , bond
order = 1
(b) N2 (11 valence electrons): ( 2s )2 (*2s )2 ( 2p )4 ( 2p )2 (*2p )1, bond
order = 2.5
(c) C2 (9 valence electrons): ( 2s )
2(*2s )2 ( 2p )4( 2p )1, bond order = 2.5
2G.14
2 * 2 4 2 * 1
2s 2s 2p 2p 2p
+ 2 * 2 4 22s 2s 2p 2p
NO (σ ) (σ ) (π ) (σ ) (π ) ;bondorder=2.5
NO (σ ) (σ ) (π ) (σ ) ;bondorder=3
Because of the higher bond order for NO+, it should form a stronger bond
and therefore have the higher bond enthalpy
2G.16 N2 and F2
have an odd number of electrons and must therefore be
paramagnetic. The molecular orbital diagram for N2 shows that adding
one electron will place it in a *2p orbital. This will give one unpaired
electron and a bond order of 2.5. The removal of an electron from 2F to
give 2F will produce one unpaired electron in the 2* p orbital O22 will be
diamagnetic, as the removal of two electrons from 2O will eliminate the
two unpaired electrons.
Focus 2 Molecules 99
2G.18 ‐ 2 * 2 4 12 2s 2s 2p 2p
2 * 2 4 ‐2 2s 2s 2p 2
‐2
(a)C (9valenceelectrons):(σ ) (σ ) (π ) (σ ) ,bondorder=2.5
C (8valenceelectrons):(σ ) (σ ) (π ) ,bondorder=2.C willhave
thestrongerbond.
(b)N (11valenceelectron 2 * 2 4 2 * 12s 2s 2p 2p 2p
2 * 2 4 22 2s 2s 2p 2p 2
s):(σ ) (σ ) (π ) (σ ) (π ) ,bondorder=2.5
N (10valenceelectrons):(σ ) (σ ) (π ) (σ ) ,bondorder=3.N willhave
thestrongerbond.
2G.20 Be2 ( 2s )2(2s
)2 ;
F2 ( 2s )2 (2s)2 ( 2 p )2 ( 2 p )4 ( 2 p
)4 ;
B2 ( 2s )
2(2s)2( 2 p )11;
C2 ( 2s )
2(2s)2( 2 p )3;
Electron affinity is the measure of the energy released when an electron is
added to an atom, molecule, or ion; the larger the electron affinity, the easier
it is to add an electron to that species. In F2 the added electron enters an
antibonding orbital, and in C2 it enters an orbital that contains another
electron. The positive charge of B2 increases its electron affinity relative to
Be2
2G.22 Given the overlap integral S A1s B1sd , the bonding orbital
A1s B1s , and the fact that the individual atomic orbitals are
normalized, we are asked to find the normalization constant N. That will
normalize the bonding orbital such that:
N 22 d N 2 (A1s B1s )2 d 1
N 2 (A1s B1s )2 d N 2 (A1s
2 2A1sB1s B1s
2 )d
N 2 A1s
2 d 2 A1sB1s d B1s
2 d
100 Focus 2 Molecules
Given the definition of the overlap integral above and the fact that the
individual orbitals are normalized, this expression simplifies to:
N 2 (1 2S 1) 1
Therefore, N 1
2 2S
To confirm that these two orbitals are orthogonal we evaluate the integral:
(A1s B1s ) (A1s + B1s ) d
A1s2 + A1s B1s A1s + B1s B1s
2 d
11 0
The overlap integral is zero. Therefore, the two wavefunctions are
orthogonal.
2.2 The Lewis structures:
The Lewis structure in the center is probably the most important as it is the
structure with the formal charges of the individual atoms closest to zero.
2.4 The charge on the Group 2 metal cations are the same, but as you go
down the group the size of the cation increases. Lattice energy
decreases with an increase in ion size; thus, as you go down the
group the lattice energy will decrease as size increases.
NC O-2 0+1
NC O-1 +1
NC O-3 +1+1-1
Focus 2 Molecules 101
2.6
2.8
2.10 There are seven equivalent resonance structures for the tropyllium cation.
All the C—C bonds will have the same bond order, which will be the
H3CC
O
O
-1H3C
CO
O-1
H3CC
O
C
-1H3C
CO
C -1HH HH
CC
C
H
H3CC
N
O
-1
H3CC
N
O-1
H
H
H
H
CC
C
H
H
H
H
H
H H
(a)
(b)
(c)
(d)
C
NC
N
CN O
O
O
N
CN
C
CC
N
H
H
HO
C
CC
N
C
C
C
H
HH
HH
H
H
H
HH C
NC
N
CN O
O
O
H
HH(a)
H H
H
H
C
N C
CN
H
H
H
H
(b) (c) (d)
102 Focus 2 Molecules
average of 4 single bonds and 3 double bonds to give an average bond
order of 1.4.
2.12 Draw Lewis structures for all four molecules:
Of these, compound (d) is the only one that cannot exist (carbon cannot
expand its octet).
2.14 In (a) and (b) all atoms have formal charges of 0;
(c) N = 0, O (single bond) = −1, O (double bond) = 0.
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C
C
C C
C
CC
H
H
H
HH
H
H
C C(a) HH C C(b)HH
H H
C C(c)HH
H H
H H
C C(d)H
HH
HH
H
H
H
Focus 2 Molecules 103
2.16
(a) Metal d(M–I), pm
(1 − d*/d)/d Lattice Energy,kJ mol1
Li 274 3.19 × 10-3 759
NaI 294 3.00 × 10-3 700
KI 329 2.72 × 10-3 645
RbI 345 2.61 × 10-3 632
CsI 361 2.51 × 10-3 601
A high correlation (R2 = 0.9731) exists between lattice energy and
d(M–I). A better fit (R2 = 0.9847) is obtained between lattice energy
and (1 − d*/d)/d.
(b) Using the equation L.E. = 218331(1 − d*/d)/d) + 54.887 and the
Ag–I distance (309 pm), the estimated AgI lattice energy is 683
kJ·mol-1.
(c) There is not close agreement between the estimated
(683 kJ·mol-1) and experimental (886 kJ·mol-1) lattice energies. A
possible explanation is that the Ag+ ion is more polarizable than the
alkali metal cations of similar size, so the bonding in AgI is more
covalent.
2.18 The potential energy well is deepest for N2 (NN), then N3– (N=N),
then N2H4 (N–N):
104 Focus 2 Molecules
2.20 (a) Mulliken defined electronegativity as 12
(IE EA); using the
values for C, N, O, and F found in Figure 2.24 and 2.28 and dividing by
230 kJ/mol, we get the following Pauling’s values (Figure 3.12) are
shown for comparison:
Element Z Mulliken
Electronegativity
Pauling
Electronegativity
C 6 2.6 2.55
N 7 3.0 3.04
O 8 3.2 3.44
F 9 4.4 3.98
(b) A plot of electronegativity versus Z for these two scales yield the
following:
Focus 2 Molecules 105
(c) The two scales are relatively close to one another, as the plot in part (b)
shows. The Pauling values seem to be a little more consistent, as a steady
rise is seen from carbon to fluorine; the Mulliken values, while showing a
general increase over the same range of Z also show a dip at oxygen and a
spike at fluorine.
2.22 (a) A reasonable Lewis structure for S6:
(b) A possible resonance structure for S6 can be drawn:
2
3
4
5
6 7 8 9
Electronegativity, MullikenElectronegativity, Pauling
Z
Ele
ctro
nega
tivity
S
SS
S
SS
S
SS
S
SS
S
SS
S
SS -1
+1
-1
-1
+1+1
S
SS
S
SS
via where indicates movement of a pair of electrons to form a double bond
106 Focus 2 Molecules
This resonance structure have higher energy than the S6 ring with no
double bonds and will therefore contribute little (if any) to the stability of
the molecule.
2.24 In each case, the more ionic pair will have the lower vapor pressure.
(a) Na2O; (b) InCl3; (c) LiH; (d) MgCl2
2.26 (a) The BrO bond in BrO− would be expected to be longer. The Br—O
bond in BrO− is a single bond, but BrO2− contains one single bond and
one double bond, so it will be shorter. (b) The SiH bond in SiH4 would
be expected to be longer. Bond distance increases with atomic size.
Silicon is larger than carbon, so it would be expected to have the longer
bond.
2.28 (a) The XX molecules will be simple diatomic molecules with an X X
single bond. The 3XX molecules will have a central atom of the VSEPR
type AX3E2. The molecule will therefore be T-shaped. The X X X
bond angles will be slightly less than 90° and 180°. The 5XX molecules
will have central atoms of the type AX5E, which will be a square
pyramidal structure with bond angles of ca. 90° and 180°. (b) All three
types will be polar. (c) The central atom is the one that is the least
electronegative. A consideration of the oxidation numbers shows that the
central atom is the one that is positive, and the atoms around it are
negative. Thus, the central atom will be the one that retains its electrons
less effectively.
2.30
C C H
H
H
sp2 spC C H
H
H
sp2 sp3H
H
C C H
H
H
sp3 sp3H
H
(a) (b) (c)
Focus 2 Molecules 107
2.32 (a) The helium and hydrogen atoms have only their 1s orbitals available
for bonding. Combination of these two will lead to a and a pair of
orbitals. The most stable species will be one in which only the orbital is
filled. Thus we need a species that has only two electrons. The charge on
this species will be +1. (b) The maximum bond order will be 1.
(c) Adding one electron will decrease the bond order by 1/2 because an
antibonding orbital will be populated. Taking an electron away will also
decrease the bond order by half because it will remove one bonding
electron.
2.34 The Lewis structures that contribute to the structure of the carbamate ion:
The charges show the location of the formal charges on the atoms in the
Lewis structure. The form that has a double bond to N is a zwitterion, a
structure that contains a positive and a negative charge in the same
molecule. We might expect this structure to have a higher energy and
contribute less to the overall structure than the other two forms, which
have less separation of charge and which are equivalent in energy. To get
insight into this question, we can compare the observed bond distances
(C—O, 128 pm, C—N, 136 pm) to those expected for various C—O and
C—N bond orders. We can estimate these values using the data given in
Table 2D.3 and Figure 2D.11. The following values are obtained:
H 1sHe 1s
1s*
1s
C O
O
NH
H
C O
O
NH
H
C O
O
NH
H
108 Focus 2 Molecules
Bond Expected Bond Length, pm
C—O 151
112, 127
C—N 152
127
The C—O bond distance is very close to what we would expect for a
, although the average of experimental data gives a value closer to
112 pm. The second value is probably more reliable indicating that the
C—O bonds in the carbamate ion are intermediate between a double and
single bond. The same is true of the C—N bond, (Experimental values of
double bonds are close to 127 pm.) It appears, then, that the
resonance form with a double bond does contribute substantially to
the structure.
2.36 The three possible forms that maintain a valence of 4 at carbon are
cyclopropene, propyne, and allene, which have these structures,
respectively:
The C—C—C bond angles in cyclopropene are restricted by the cyclic
structure to be approximately 60°, which is far from the ideal value of
109.47° for an sp3-hybridized carbon atom, or 120° for an sp2-hybridized
C atom. Consequently, cyclopropene is a very strained molecule that is
extremely unstable. The H—C—H bond angle at the CH2 group would be
expected to be 109.47°, but it is actually larger because of the narrow
C—C—C bond angle. The carbon of the CH2 group is sp3-hybridized,
whereas the other two carbon atoms are sp2. Likewise the C=C—H angles,
which would normally be expected to be 120° because of sp2 hybridization
at C, are somewhat larger. In allene, the middle carbon atom is sp-
hybridized; the two end carbons are sp2-hybridized. The C—C—C bond
C
C
CH
H
H
H
H C
H
H
C C H C C C
H
H
H
H
Focus 2 Molecules 109
angle is expected to be 180°; the H—C—H bond angles should be close to
120°. In propyne, one end carbon is sp3-hybridized (109.5° angles); the
other two C atoms are sp-hybridized, with 180° bond angles. The three
structures are not resonance structures of each other, as they have a
different spatial arrangement of atoms. For two structures to be resonance
forms of each other, only the positions of the electrons may be changed.
These two compounds are isomers.
2.38 To show that two orbitals taken together are cylindrically symmetric, we
assume that the z-axis is the bonding axis and show that net probability
distribution (the sum of the two molecular orbitals squared) is not a
function of . (Refer to Figure 2.2 to confirm that, looking down the z-
axis, a cylindrically symmetric orbital will not be a function of .)
x f (z) f (z) C sin cosy f (z) f (z) C sin sin
where C 3
4
12
, and f (z) is not a function of x or y (and, therefore,
f (z) is not a function of ). Squaring the two wavefunctions and summing them:
f (z)2 C 2 sin2 cos2 f (z)2 C 2 sin2 sin2 f (z)2 C2 sin2 cos2 sin2 Using the identity cos2 x sin2 x 1 this becomes f (z)2 C 2 sin2With two electrons in each orbital, the electron distribution is nota function of and is, therefore, cylindrically symmetric about thebonding axis.
2.40 The overlap can be end-to-end or side-on. In the side-on overlap situation,
the net overlap is zero because the areas of the wave function that are
negative will cancel with the areas of the wave function that are positive.
These will be equal in area and opposite in sign, giving no net overlap.
110 Focus 2 Molecules
2.42
For the trigonal planar molecule, we first construct the vectors
representing the individual dipole moments and then add them. This can
be done most easily by combining two of the vectors first and then adding
that to the third. Let’s first add b and c. In order to add the vectors, we
need to position them head to tail. Because the original angle where b and
c come together in the AX3 molecule is 120, the angle between them
when the origin of c is shifted to the end of b will be 60. Because that
angle is 60 and b is the same length as c, the resultant of b + c must also
have the same length as a (and b and c). As can be seen from the diagram,
the original vector a points in exactly the opposite direction from the
summed vector b + c. Because the angle between b and b + c is 60 and
the angle between a and b is 120, it follows that the angle between a and
b + c must be180. Because b + c has exactly the same magnitude as a,
they will exactly cancel each other.
(a) (b)
nonbondingbonding
a
b
c
(c)
(b+c)
a
b
c
Focus 2 Molecules 111
For the tetrahedron, which can be inscribed inside a cube as shown above,
the vectors can be added in pairs to give resultants as shown. Starting with
original vectors a, b, c, and d, we can add a + b and c + d. (For an ideal
tetrahedron in which all the atoms bonded to the central atom are identical,
the choice of which vector is a, b, c, or d is arbitrary because they are all
equivalent.) An examination of the tetrahedron will show that the vectors a
and b lie in a plane perpendicular to the plane of vectors c and d. The
resultant c + d will be equal in magnitude to the resultant a + b because the
vectors a, b, c, and d all have the same magnitude, and the angle between
them is the same. It is easy to see that the vectors a + b and c + d lie exactly
opposite each other and will add to zero.
2.44 (a) The structure of caffeine and the hybridization of each nonhydrogen atom
are shown below. For the sake of clarity all lone pairs of electrons have
been left out; each N in the structure has one lone pair not shown, while
each O has two lone pairs not shown:
(b) The bond angles around any atom marked sp3should be 109.5 while the
bond angles around any atom marked sp2 should be 120.
OC
N
CH3
C
CN
CH3
C
N
NCH3 C
O
H
sp3
sp3
sp3
sp3
sp3 sp2
sp2sp2
sp2
sp2sp2
sp2
sp2
sp3
112 Focus 2 Molecules
(c) The crystal structure of caffeine is known. See "The Structures of the
Pyrimidines and Purines. VII. The Crystal Structure of Caffeine." D. June
Sutor. Acta Cryst. (1958). 11, 453-458 for bond length and angles. The
experimental values compare favorably with those predicted.
2.46 (a)
(b) The hybridization at the atoms attached to two hydrogen atoms is sp2,
whereas that at the carbon atoms attached only to two other carbon atoms
is sp.
(c) Double bonds connect all of the carbon atoms to each other.
(d) The H—C—H and C—C—H angles should all be ca. 120°. The
C—C—C angles will all be 180°.
(e) The hydrogen atoms in H2CCH2 and H2CCCCH2 lie in the same plane,
whereas the planes that are defined by the two end CH2 groups lie
perpendicular to each other in H2CCCH2. This is because the p orbitals
that are used by the central carbon atom to form the double bonds to the
end carbon atoms are perpendicular to each other as shown. Diagram of
the interaction of the p orbitals used in making the C—C double bonds:
Diagram showing the orientation of the sp2 orbitals used for bonding to the
H atoms:
C C C
H
H
H
H
C C C
H
H
C C
H
H
C
H
H
H
H
C C C
px - pxoverlap
py - pyoverlap
90°
Focus 2 Molecules 113
The three different p-orbitals are commonly labeled px, py, and pz to
distinguish them. They may be thought of as pointing in the x, y, and z
directions on a set of Cartesian axes that intersect at the carbon atom. The
three p-orbitals on a given carbon atom will be oriented 90 apart. Thus,
if the central carbon atom is sp-hybridized, the sp-hybrid orbitals will be
180 apart (not shown) and will form the bonds to the other carbon
atoms. This arrangement will use the p-orbitals on the end carbon atoms
that are oriented in the same direction. (As shown here, the pz-orbitals on
all three carbons have been used.) The px- and py-orbitals on the middle
carbon will be used in forming the bonds. This means that the bond to one
carbon atom will be made from px, px interactions and the bond to the
other will be py, py. One end carbon will be using the s-, pz-, and px-
orbitals for forming the sp2-hybrids and the other will be using the s-, pz-,
and py-orbitals. This change will result in the sp2-orbitals on the end
carbons being oriented 90away from each other. This orientation effect
will occur only if there is an odd number of carbons in the chain. If there
is an even number of carbon atoms, the end C atoms will be required to
use the same type of p-orbitals as each other for forming the sp2-hybrid
orbitals. Thus H2CCCCH2 should have all the hydrogen atoms in the
same plane, and in H2CCCCCH2 the planes of the end CH2 groups will
again be perpendicular to each other.
(f) If x is odd, the hydrogen atoms on one end will lie in a plane
perpendicular to the plane occupied by the hydrogen atoms on the other
end. If x is even, all of the hydrogen atoms will be coplanar.
2.48 (a) The hybridization for each C and N atom is as follows:
C C CH
H
H
H
90°
114 Focus 2 Molecules
(b) 1 has 16 bonds and 4 bonds; 2 has 15 bonds and 4 bonds.
(c) There are 5 lone pairs of electrons in 1 and 6 lone pairs in 2.
2.50 The germide ion Ge4n–has the following structure:
This structure has a total of 20 electrons used as either lone or bonding
pairs; four Ge atoms can provide only 16 of these electrons (4 Ge atoms
at 4 valence electrons per Ge), meaning there must be four extra
electrons. Therefore –n –4.
2.52 Other answers are also possible; to be isoelectronic both species only
need to have the same number of total valence electrons.
(a) BF3 and CO32 are isoelectronic (24 valence electrons); both are
trigonal planar.
(b) SO2 and O3 are isoelectronic (18 valence electrons); both are bent
(with central atom bond angles of 120).
(c) HBr and OH– are isoelectronic (8 valence electrons); both are linear.
HC
NC
CN
C
NH
NC
NH2
H
sp3
sp2
sp2
sp3
1
sp2
sp2sp2sp2
sp2
sp2
sp2
HC
NC
CN
C
NH
NH C
O
H
2
sp3
sp2
sp2
sp3sp2
sp2sp2sp2
sp2
Ge
GeGe
Gen-
Focus 2 Molecules 115
2.54
(b) The bond order between the 2 nitrogens is 1. (c) Nitrogen is unable
to have an expanded octet, whereas phosphorus can have an expanded
octet. Thus you will see different structures between the two.
2.56 (a) The Lewis structures:
Both phosphite and nitrite have standard Lewis structures; upon oxidation
each picks up one more oxygen. Phosphate is able to expand its octet to
accommodate 10 electrons around it while nitrate cannot. In doing so
phosphate can eliminate both a positive and a negative formal charge, thus
lowering its overall energy:
This ability of P to expand its octet is due to the presence of unoccupied 3d
ortbitals. Nitrogen, which is a Period 2 element does not have d orbitals and
therefore cannot expand its octet.
N
O
N
O
O
(a)
O P O
O-1
-1
-1
3
O P O
O
O
-1
-1
-1 3-
O
N OONO
O-1 -1
+1
-1
nitrate nitrite
phosphate phosphite
O P O
O
O
-1
-1
-1 3-
O P O
O
O
-1
-1
-1 3-
-1+1
116 Focus 2 Molecules
(b) Phosphate ions can form chains in which one phosphate can be linked
to another (polyphosphates or phosphate anhydrides):
Nitrate, on the other hand, cannot form such species; presumably, this is
because formation of a polynitrate would require placing a number of
positive charges relatively close to one another, an unfavorable
arrangement:
2.58
2.60
O
P
O
O O
O
P
O
O
O
P
O
O
-1 -1 -1
N
O
O O N
O
O N
O
O+1 +1 +1
H3CN
CH3
N
O
Cl C Cl
O
N C N
O
NN
N
N
O
O
O
O
O
O-1
-1
-1
+1
+1+1
H
HH
H
(a) (b)
(c)
Focus 2 Molecules 117
2.62 (a) The angles represented by a and b are expected to be about 120while c
is expected to be about 109.5 in 2, 4-pentanedione. All of the angles are
expected to be about 120 in the acetylacetonate ion.
(b) The major difference arises at the C of the original sp3-hybridized CH2
group, which upon deprotonation and resonance goes to sp2 hybridization
with only three groups attached (the double-headed arrow is read as “in
resonance with”):
2.64 The Lewis structure of hydroxylamine:
The hybridization of the both the oxygen and the nitrogen is sp3. All bonds
in hydroxylamine are -bonds; there are no -bonds present. Because of
the presence of the lone pairs on oxygen and nitrogen, both the H—O—N
and the O—N—H angles are less than their predicted 109.5°.
2.66 (a) The Lewis structure of hydrogen peroxide is
No atoms have formal charges, and the oxidation number of each O is –1.
Oxidation number is more useful than formal charge in predicting the
oxidizing ability of a compound, as this will give us a clearer idea of
whether a given atom wants to gain or lose an electron.
(b) The bond angles should be about the same as that for water, ca. 104°.
All atoms in this molecule are in the same plane, but because of
unrestricted rotation around the O—O bond, it is expected that the
hydrogens (and therefore the lone pairs of electrons on the oxygens) can
H3CC
CC
CH3
O O
H
H3CC
CC
CH3
O O
H
OH N H
H
H O O H
118 Focus 2 Molecules
rotate in and out of the plane; one of these rotations will lead to a polar
structure. (The different forms, arising from rotations around a single
bond, are called rotomers.) Because of this, the molecule is expected to be
polar, but slightly less so than water.
(c) (1) O2: ( 2s )2 (2s)2 ( 2 p )2 ( 2 p )4 (2 p
)2; bond order = 2;
paramagnetic.
(2) O2
: ( 2s )2(2s
)2( 2 p )2 ( 2 px)4 (2 px
)3; bond order = 1.5;
paramagnetic.
(3) O2: ( 2s )
2(2s)2( 2 p )2 ( 2 p )4(2 p
)1; bond order = 2.5;
paramagnetic.
(4) O22: ( 2s )
2(2s)2( 2 p )2 ( 2 p )4(2 p
)4; bond order = 1; diamagnetic.
(d) Bond length should inversely follow bond order; therefore, the longest
bond should belong to the molecule with the lowest bond order (O22 ), and
the shortest bond should belong to the species with the largest bond order
(O2 ) , which is borne out in the data.
(e) 2 Fe2+ + H2O2 + 2 H+ 2 Fe3++ 2 H2O
‐32 22 2 2 2
2+ 3+ 3+‐3 3+
2 2 2+ 3+2 2
0.200molH O1L43.2mLH O × × =8.64 10 molH O
1000mL 1L
2molFe 2molFe 55.85gFe8.64 10 molH O =0.965gFe
1molH O 2molFe 1molFe
(f) 2 Fe3+ + H2O2 + 2 OH- 2 Fe2+ + 2 H2O + O2
HO
O
H
HO
O
H
planar, nonpolarrotomer
planar, polarrotomer
HO
O H
nonplanar, polarrotomer (one of many)
Focus 2 Molecules 119
32 22 2 2 2
3 2 23 3
2 2 3 22 2
0.200molH O1L41.8mLH O 8.36 10 molH O
1000mL 1L
2molFe 2molFe 55.85gFe8.36 10 molH O 0.934gFe
1molH O 2molFe 1molFe
(g) Two electrons are transferred.
(h) The homolytic cleavage of hydrogen peroxide to form hydroxyl
radicals is:
The energy required to break this bond is 157 kJ/mol. To break a single
O—O will take:
Eh
2.611019 J6.626 1034 J sec
3.94 1014 Hz
c
2.998 108 m s1
3.94 1014 s1 7.61107 m = 761 nm
H O O H H O O H+
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